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Official PYQ · 2014Set 3 · Q5Linear Algebra: Matrices, Determinants, System of Linear EquationsNAT1 mark

GATE CS 2014 Set 3 Q5 — Linear Algebra: Matrices, Determinants, System of Linear Equations

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If V1V_1 and V2V_2 are 44-dimensional subspaces of a 66-dimensional vector space VV, then the smallest possible dimension of V1∩V2V_1 \cap V_2 is _____.
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Answer: 2

We are given that V1V_1 and V2V_2 are 44-dimensional subspaces of a 66-dimensional vector space VV. Therefore:

  • dim⁡(V)=6\dim(V) = 6
  • dim⁡(V1)=4\dim(V_1) = 4
  • dim⁡(V2)=4\dim(V_2) = 4

By the subspace dimension theorem, the relation between the dimensions of the sum and intersection of two subspaces is given by:

dim⁡(V1+V2)=dim⁡(V1)+dim⁡(V2)−dim⁡(V1∩V2)\dim(V_1 + V_2) = \dim(V_1) + \dim(V_2) - \dim(V_1 \cap V_2)

Rearranging the formula to solve for the dimension of the intersection:

dim⁡(V1∩V2)=dim⁡(V1)+dim⁡(V2)−dim⁡(V1+V2)\dim(V_1 \cap V_2) = \dim(V_1) + \dim(V_2) - \dim(V_1 + V_2)dim⁡(V1∩V2)=4+4−dim⁡(V1+V2)=8−dim⁡(V1+V2)\dim(V_1 \cap V_2) = 4 + 4 - \dim(V_1 + V_2) = 8 - \dim(V_1 + V_2)

To find the smallest possible dimension of V1∩V2V_1 \cap V_2, we need to maximize the dimension of V1+V2V_1 + V_2.

Since V1+V2V_1 + V_2 is a subspace of VV, its dimension cannot exceed the dimension of the parent vector space VV:

dim⁡(V1+V2)≤dim⁡(V)=6\dim(V_1 + V_2) \le \dim(V) = 6

Substituting the maximum possible value of dim⁡(V1+V2)=6\dim(V_1 + V_2) = 6 into our equation:

Minimum dim⁡(V1∩V2)=8−6=2\text{Minimum } \dim(V_1 \cap V_2) = 8 - 6 = 2

Thus, the smallest possible dimension of V1∩V2V_1 \cap V_2 is 2.

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